An apple fell on Newton's head. Not yours! Good. Have you not seen a book or notebook falling from the table? I have seen full tray overturning from the table, as soon as a glass has been removed.
Let us perform the experiment of falling from the tables; of course not of the book or notebook. Take a hard thin rectangular cardboard and place it on the table.
Slowly move it towards the edge of the table, and beyond it, not allowing it to fall. Be very slow in moving more and more of it beyond the edge. Can you bring it to a situation that even the slightest push would make it fall! With your pencil, mark a line on the cardboard along the edge of the table.
The line divides the weight of the cardboard in two equal parts. Can you find more such lines on the cardboard? Do they intersect? (Can there be such lines not intersecting?) Mark the point of intersection. Sharpen your pencil pin-pointed or else you may take a pin. Could you balance the cardboard on your pin at this point?
Now take cardboard of different shapes, regular and irregular.
Could you discover balancing points in each of them? Enough of balancing act, you qualify to be a joker in my circus!!! Ha Ha.
Let us take the triangular cardboard $ABC$. Keeping $A$ on the edge of the table, keep $B$ and a little more off the table. Without moving $A$, take $B$ farther and farther from the table; in search of the point $D$ on $BC$, so that $AD$ becomes the dividing line for the weight of the triangle $ABC$. Can you guess the location of $D$, so that $AD$ is dividing the area $ABC$ in two equal halves? Simple $\ldots$ height-base formula for the area ensures that $D$ must be the mid-point. You actively knew that $AD$, the median is the dividing line. $BE$, the other median is the other dividing line. The point $P$; the point of intersection of $AD$ and $BE$ must be the balancing point for the triangle $ABC$. The balancing point could be obtained by taking the intersection of medians through $A$ and $C$ also. Can there be more than one balancing point to triangle $ABC$? If not, physically at least the third median must meet the other two at the same point $P$. What do you discover, write it out and prove it analytically.
Discover the balancing points in Cardboards of following shapes.
Discuss your achievements and your problems.
Discover balancing points in the three dimensional objects. Discuss whether a glass full of milk is more stable or when emptied. Make a glass taller, it becomes more stable or less, discuss.
Let us perform the experiment of falling from the tables; of course not of the book or notebook. Take a hard thin rectangular cardboard and place it on the table.
Slowly move it towards the edge of the table, and beyond it, not allowing it to fall. Be very slow in moving more and more of it beyond the edge. Can you bring it to a situation that even the slightest push would make it fall! With your pencil, mark a line on the cardboard along the edge of the table.
What is the significance of this line to this cardboard piece?
The line divides the weight of the cardboard in two equal parts. Can you find more such lines on the cardboard? Do they intersect? (Can there be such lines not intersecting?) Mark the point of intersection. Sharpen your pencil pin-pointed or else you may take a pin. Could you balance the cardboard on your pin at this point?
Now take cardboard of different shapes, regular and irregular.
Could you discover balancing points in each of them? Enough of balancing act, you qualify to be a joker in my circus!!! Ha Ha.
Let us take the triangular cardboard $ABC$. Keeping $A$ on the edge of the table, keep $B$ and a little more off the table. Without moving $A$, take $B$ farther and farther from the table; in search of the point $D$ on $BC$, so that $AD$ becomes the dividing line for the weight of the triangle $ABC$. Can you guess the location of $D$, so that $AD$ is dividing the area $ABC$ in two equal halves? Simple $\ldots$ height-base formula for the area ensures that $D$ must be the mid-point. You actively knew that $AD$, the median is the dividing line. $BE$, the other median is the other dividing line. The point $P$; the point of intersection of $AD$ and $BE$ must be the balancing point for the triangle $ABC$. The balancing point could be obtained by taking the intersection of medians through $A$ and $C$ also. Can there be more than one balancing point to triangle $ABC$? If not, physically at least the third median must meet the other two at the same point $P$. What do you discover, write it out and prove it analytically.
Discover the balancing points in Cardboards of following shapes.
Discover balancing points in the three dimensional objects. Discuss whether a glass full of milk is more stable or when emptied. Make a glass taller, it becomes more stable or less, discuss.
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